$A$ cell is connected to a potentiometer, and the balance point is obtained at a length of $2 \, m$. When a resistance of $5 \, \Omega$ is connected in parallel with the cell, the balance point is obtained at a length of $3 \, m$. What is the internal resistance of the cell in $\Omega$?

  • A
    $1.5$
  • B
    $10$
  • C
    $15$
  • D
    $1$

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Similar Questions

In an experiment of a potentiometer for measuring the internal resistance of a primary cell,a balancing length $\ell$ is obtained on the potentiometer wire when the cell is in an open circuit. Now,the cell is short-circuited by a resistance $R$. If $R$ is equal to the internal resistance of the cell,the balancing length on the potentiometer wire will be:

In a potentiometer experiment,the null point is obtained on the $7^{\text{th}}$ wire for a given cell. To shift the null point to the $9^{\text{th}}$ wire for the same cell,what should we do?

$A$ potentiometer is used for the comparison of $e.m.f.$ of two cells $E_1$ and $E_2$. For cell $E_1$,the null point is obtained at $20 \ cm$ and for $E_2$,the null point is obtained at $30 \ cm$. The ratio of their $e.m.f.$s will be:

In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$ and the balancing length is $80 \ cm$. If the polarity of $E_2$ is reversed,the balancing length becomes $20 \ cm$. The ratio $E_1 / E_2$ is:

With the help of a potentiometer,we can determine the value of the emf of a given cell. The sensitivity of the potentiometer is:
$(A)$ directly proportional to the length of the potentiometer wire
$(B)$ directly proportional to the potential gradient of the wire
$(C)$ inversely proportional to the potential gradient of the wire
$(D)$ inversely proportional to the length of the potentiometer wire
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